Question: Divide the following complex numbers. $\dfrac{25+19i}{5-3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${5+3i}$. $ \dfrac{25+19i}{5-3i} = \dfrac{25+19i}{5-3i} \cdot \dfrac{{5+3i}}{{5+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(25+19i) \cdot (5+3i)} {5^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(25+19i) \cdot (5+3i)} {(5)^2 - (-3i)^2} $ $ = \dfrac{(25+19i) \cdot (5+3i)} {25 + 9} $ $ = \dfrac{(25+19i) \cdot (5+3i)} {34} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({25+19i}) \cdot ({5+3i})} {34} $ $ = \dfrac{{25} \cdot {5} + {19} \cdot {5 i} + {25} \cdot {3 i} + {19} \cdot {3 i^2}} {34} $ $ = \dfrac{125 + 95i + 75i + 57 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{125 + 95i + 75i - 57} {34} = \dfrac{68 + 170i} {34} = 2+5i $